Equinormal solutions of two weak acids `HA(pK_(a)=3)` and `HB(pK_(a)=5)` are each placed in contact with standard hydrogen electrode at `25^(@)C(T=298K)` when a cell is constructed by interconnecting them through a salt bridge find the e.m.f. of the cell.

Equinormal solutions of two weak acids `HA(pK_(a)=3)` and `HB(pK_(a)=5

1.	Equinormal solutions of two weak acids `HA(pK_(a)=3)` and `HB(pK_(a)=5)` are each placed in contact with standard hydrogen electrode at `25^(@)C(T=298K)` when a cell is constructed by interconnecting them through a salt bridge find the e.m.f. of the cell.
Answer» Correct Answer - `E=0.059V` `Pt//H_(2)^(@)//H^(+)(HA)//H^(+)(HB)//H_(2)//Pt` `H_(2)to2H^(+)+2e^(-)` `2H^(+)+2e^(-)toH_(2)` `2H_(HB)^(+)to2H_(HA)^(+)" "E^(@)=0` `E=0-(0.0591)/(2)log(([H^(+)]_(HA)^(2))/([H^(+)]_(HB)^(2)))` But `Ka=([H^(+)]^(2))/(c)` `=-(0.0591)/(2)log((10^(-3)xxC)/(10^(-5)xxC))=-0.0591` the cell is constructed in reversed direction. `E_(cell)=0.0591"volt"`

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Equinormal solutions of two weak acids `HA(pK_(a)=3)` and `HB(pK_(a)=5)` are each placed in contact with standard hydrogen electrode at `25^(@)C(T=298K)` when a cell is constructed by interconnecting them through a salt bridge find the e.m.f. of the cell.

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